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For what negative value of k is there exactly one solution to the system of equations \(\begin{align*} y &= 2x^2 + kx + 6 \\ y &= -x + 4? \end{align*}\)

 

The quadratic x^2-3x+1 can be written in the form (x+b)^2+c, where b and c are constants. What is b+c?

 

A parabola with equation y=x^2+bx+c passes through the points (-1,-11) and (3,17). What is c?

 

What is the value of c if \(x\cdot(3x+1)<c\) if and only when \(x\in \left(-\frac{7}{3},2\right)\)?

 Jul 16, 2016

Best Answer 

 #1
avatar+118658 
+5

It is best to ask just one question at a time.

Learn from the answer or ask questions about it.

THEN

move to the next question if you still need help with it.

 

\(\begin{align*} y &= 2x^2 + kx + 6 \\ y &= -x + 4? \end{align*}\\~\\ y=2x^2+kx+6\\ -x+4=2x^2+kx+6\\ 0=2x^2+kx+x+6-4\\ 0=2x^2+(k+1)x+2\\ \mbox{If there is to be only one solution, the discriminant must equal 0}\\ \triangle=(k+1)^2-16=0\\ (k+1)^2=16\\ k+1=\pm4\\ \mbox{The only negative solution for k is }\\ k=-5 \)

 

 

Here is the graph.  

https://www.desmos.com/calculator/29trjct5fv

 Jul 17, 2016
 #1
avatar+118658 
+5
Best Answer

It is best to ask just one question at a time.

Learn from the answer or ask questions about it.

THEN

move to the next question if you still need help with it.

 

\(\begin{align*} y &= 2x^2 + kx + 6 \\ y &= -x + 4? \end{align*}\\~\\ y=2x^2+kx+6\\ -x+4=2x^2+kx+6\\ 0=2x^2+kx+x+6-4\\ 0=2x^2+(k+1)x+2\\ \mbox{If there is to be only one solution, the discriminant must equal 0}\\ \triangle=(k+1)^2-16=0\\ (k+1)^2=16\\ k+1=\pm4\\ \mbox{The only negative solution for k is }\\ k=-5 \)

 

 

Here is the graph.  

https://www.desmos.com/calculator/29trjct5fv

Melody Jul 17, 2016
 #2
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+5

y=x2+bx+c, passes through (-1, -11) and (3, 17). c=?

A. 11=1-b+c

B. 17=9+3b+c

C. c=b-12

D. 17=4b-3->4b=20

b=5

c=-7

 

x2-3x+1=(x+b)2+c

b+c=?

(x+b)2+c=x2+2bx+b2+c=x2-3x+1

2bx=-3x

b=-3/2

b2+c=1

c=1-9/2

c=-7/2

b+c=-3/2+-7/2=-10/2=-5

 Aug 4, 2016
 #3
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0

What is the value of c if x(3x+1)<c only when x is (-7/3, 2)

3x2+x<c

3x2+x-c<0

3=a, 1=b, -c=c, x=(-b+\sqrt(b2-4ac))/2a

x<(-1+\sqrt(1+12c))/6

Test: x=-7/3

-14<-1+\sqrt(1+12c)

-13<+\sqrt1+12c

169>1+12c

168>12c

c<14

Test: x=2

12<-1+\sqrt(1+12c)

13<+\sqrt(1+12c)

169<1+12c

168<12c

c>14

 

If c<14 and c>14, then c=14

 Aug 4, 2016
 #4
avatar+129840 
0

What is the value of c if  x * (3x + 1) < c    and  x   lies on the interval (-7/3, 2)

 

This is a parabola that turns upward

 

At x = -7/3,  x (3x + 1 )   = 14

 

At x = 2, x (3x + 1)  = 14

 

So.....on the interval  (-7/3, 2), because of parabolic symmetry, all f(x)  will be < 14....thus...   the inequality will be true  whenever c = 14

 

 

 

cool cool cool

 Aug 4, 2016

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